At which speed and angle should satellite be to orbit in a longitude line (on Greenwich line, for example) ? most satellites I know move in latitude lines (equator line), or just a part of longitude line
The only satellites that move on latitude lines are above the equator, ie at 0 latitude. And no satellites move on longitude lines. It just isn't possible.
The vast majority of satellites have roughly sinusoidal orbital tracks on the surface of the Earth. Even polar orbiting satellites do not have a track that follows a line of longitude.
The closest to what you are looking for is probably a Molniya orbit, which can spend a lot of its time closely aligned with a line of longitude. See this picture from Wikipedia:
If the Earth didn't spin (or spins very very slowly), you certainly have found the answer: i=90°, i.e. a perfect polar orbit.
Since the Earth spins, the ground-track of a polar orbit can't follow the (moving) longitude. Imagine the ground-track crosses the Equator at the Greenwich meridian at a certain time t0. When the satellite reaches the latitude of Greenwich at time t>t0, this town would have moved Eastward in the mean time. You can play with the online Ground track visualizer to see this effect (the groundtrack is not perpendicular to the Equator).
Nevertheless, if you decide to ignore the high-latitude areas (North or South), then for a given satellite altitude, you can select an inclination slightly lower than 90° so that the groundtrack of the satellite appears to "hug" the longitudes, up to the chosen limiting Latitudes. That is, the "non-perfect" polar orbit gets compensated somehow, resulting in the groundtrack appearing as perpendicular to the Equator, for most part of the orbit revolution. As a numerical example, at ~1200 Km, the corresponding inclination is ~86°, up to +/- 60° Latitude (where most people live). Incidentally, these are the parameters selected by the Oneweb constellation. Hence, if a Oneweb satellite crosses the Equator at 0° Longitude, it will be at Greenwich zenith when it crosses Greenwich's latitude (approximately).